Question

A vessel contains liquids P and Q in the ratio 5 : 3. If 16 L of the mixture is removed and replaced by same quantity of liquid Q and the ratio becomes 3:5 What is the quantity that the vessel hold ?

• A. 35L
• B. 40L
• C. 45L
• D. 50L

Answer 1

Answer:

B. 40L

Initial State: The vessel initially contains liquids P and Q in the ratio 5:3. This means that for every 5 parts of P, there are 3 parts of Q. If the total quantity in the vessel is 'V' liters, then the initial quantity of P is (5/8)V and Q is (3/8)V.

• Removal of Mixture: When a certain quantity of the mixture is removed, the ratio of the components in the remaining mixture does not change. For example, if 16 L of the mixture is removed:

  • Quantity of P removed = (5/8) × 16 L = 10 L

  • Quantity of Q removed = (3/8) × 16 L = 6 L

  • The remaining quantities are (5/8)V - 10 L of P and (3/8)V - 6 L of Q. The ratio of P:Q in the remaining mixture is still 5:3.

• Replacement: When 16 L of liquid Q is added back, the total volume in the vessel returns to its original capacity, 'V'. However, the quantities of individual components change:

  • The quantity of P remains (5/8)V - 10 L (as no P was added).

  • The quantity of Q becomes (3/8)V - 6 L (remaining Q) + 16 L (added Q) = (3/8)V + 10 L.

• New Ratio and Equation Setup: The problem states that the new ratio of P:Q becomes 3:5. Therefore, we can set up an equation: (Quantity of P) / (Quantity of Q) = 3/5.

  • [(5/8)V - 10] / [(3/8)V + 10] = 3/5

  • Cross-multiply: 5 × [(5/8)V - 10] = 3 × [(3/8)V + 10]

  • (25/8)V - 50 = (9/8)V + 30

• Solving for Total Quantity (V):

  • Bring terms with 'V' to one side and constants to the other:

  • (25/8)V - (9/8)V = 30 + 50

  • (16/8)V = 80

  • 2V = 80

  • V = 40 L

  Therefore, the quantity that the vessel can hold is 40 L.